Answer
$x=8$
Work Step by Step
$\sqrt{3x+1}=2+\sqrt{x+1}$
Square both sides of the equation:
$(\sqrt{3x+1})^{2}=(2+\sqrt{x+1})^{2}$
$3x+1=4+4\sqrt{x+1}+x+1$
$3x+1=x+5+4\sqrt{x+1}$
Reorganize the equation:
$3x+1-x-5=4\sqrt{x+1}$
$2x-4=4\sqrt{x+1}$
Once again, square both sides of the equation:
$(2x-4)^{2}=(4\sqrt{x+1})^{2}$
$4x^{2}-16x+16=16(x+1)$
$4x^{2}-16x+16=16x+16$
Take all terms to the left side:
$4x^{2}-16x+16-16x-16=0$
$4x^{2}-32x=0$
Solve this equation by factoring. Take out common factor $4x$:
$4x(x-8)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$4x=0$
$x=\dfrac{0}{4}$
$x=0$
$x-8=0$
$x=8$
Check the solutions found by plugging them into the original equation:
$x=0$
$\sqrt{3(0)+1}=2+\sqrt{0+1}$
$\sqrt{1}=2+\sqrt{1}$
$1\ne3$ False
$x=8$
$\sqrt{3(8)+1}=2+\sqrt{8+1}$
$\sqrt{25}=2+\sqrt{9}$
$5=2+3$
$5=5$ True
The final answer is $x=8$
